Proof of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval

Claim 1 is exactly Continuous Functions on Compact Intervals are Riemann Integrable, whose hypothesis that $h$ be continuous on $[a,b]$ is the continuity at every point of $[a,b]$ assumed in the statement. Throughout we use that the metric of The Absolute Value Metric on the Real Line is $d_{\mathbb{R}}(s,t)=|s-t|$, so that continuity of $h$ at $x$ relative to $[a,b]$ says: for every $\varepsilon>0$ there is $\delta>0$ such that every $y\in[a,b]$ with $|y-x|<\delta$ satisfies $|h(y)-h(x)|<\varepsilon$. Write $I=\int_a^bh(x)\,dx$ for the Riemann integral of Riemann Integrability on a Closed Interval: for every $\varepsilon>0$ there is $\delta>0$ such that every Riemann sum of $h$ over a tagged partition of mesh less than $\delta$ differs from $I$ by less than $\varepsilon$.

Step 1 (measurability and integrability). Fix $t\in\mathbb{R}$. For each $x\in[a,b]$ with $h(x)>t$, continuity of $h$ at $x$ relative to $[a,b]$, applied with $\varepsilon=h(x)-t$, gives $\delta_x>0$ such that $h>t$ on $(x-\delta_x,x+\delta_x)\cap[a,b]$; with $G=\bigcup_x(x-\delta_x,x+\delta_x)$ (union over such $x$), an open set, $\{x\in[a,b]:h(x)>t\}=G\cap[a,b]$. Hence $\{x\in\mathbb{R}:\tilde{h}(x)>t\}$ equals $G\cap[a,b]$ if $t\ge0$ and $(G\cap[a,b])\cup(\mathbb{R}\setminus[a,b])$ if $t<0$; in both cases a Borel set, since open sets and closed sets are Borel and $\mathcal{B}(\mathbb{R})$ is closed under finite intersections and unions. As $t\in\mathbb{R}$ was arbitrary, claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line shows that $\tilde{h}$ is Borel measurable. The same applies to $-\tilde{h}$, which is the zero extension of $-h$ and so satisfies the same hypotheses; since $\{|\tilde{h}|>t\}=\{\tilde{h}>t\}\cup\{-\tilde{h}>t\}$ for $t\ge0$ and $\{|\tilde{h}|>t\}=\mathbb{R}$ for $t<0$, claim 3 of that lemma also gives the measurability of $|\tilde{h}|=\tilde{h}^{+}+\tilde{h}^{-}$ (positive and negative parts as in Integrable Function and the Lebesgue Integral). By Extreme Value Theorem on a Closed Real Interval, applicable because $h$ is continuous on $[a,b]$ as a map into $(\mathbb{R},d_{\mathbb{R}})$, there are $x_{\min},x_{\max}\in[a,b]$ with $h(x_{\min})\le h(x)\le h(x_{\max})$ for every $x\in[a,b]$. Let $M$ be the larger of $|h(x_{\min})|$ and $|h(x_{\max})|$, a finite nonnegative real number. For $x\in[a,b]$ we get $h(x)\le h(x_{\max})\le M$ and $-h(x)\le-h(x_{\min})\le M$, so $|h|\le M$ on $[a,b]$ and $|\tilde{h}|\le M\mathbf{1}_{[a,b]}$ pointwise. Since the simple function $M\mathbf{1}_{[a,b]}$ has integral $M\,\lambda([a,b])=M(b-a)$ — Lebesgue measure assigns each interval its length — monotonicity (Linearity and Monotonicity of the Lebesgue Integral) gives $\int_{\mathbb{R}}|\tilde{h}|\,d\lambda\le M(b-a)<\infty$, so $\tilde{h}$ is integrable. This proves Claim 2.

Step 2 (reduction to a nonnegative integrand). Let $k=h+M$. Then $k\ge0$ on $[a,b]$ (since $h\ge-|h|\ge-M$ there), and $k$ is continuous on $[a,b]$, since $|k(y)-k(x)|=|h(y)-h(x)|$ for all $x,y\in[a,b]$, so that any $\delta$ witnessing the continuity of $h$ at $x$ witnesses that of $k$. On the Riemann side we work directly with Riemann Integrability on a Closed Interval: every Riemann sum of $k$ over a tagged partition equals the corresponding Riemann sum of $h$ plus $M(b-a)$, because $\sum_i(h(\xi_i)+M)(t_i-t_{i-1})=\sum_ih(\xi_i)(t_i-t_{i-1})+M(b-a)$; hence (with the same $\delta$ for each $\varepsilon$) $k$ is Riemann integrable with $\int_a^bk(x)\,dx=I+M(b-a)$. On the Lebesgue side, the zero extension of $k$ is $\tilde{h}+M\mathbf{1}_{[a,b]}$, whose integral is $\int\tilde{h}\,d\lambda+M(b-a)$ by linearity for integrable functions (Linearity and Monotonicity of the Lebesgue Integral). So Claim 3 for $k$ implies Claim 3 for $h$, and we may assume $h\ge0$.

Step 3 (sandwich, valid for every partition). Let $\mathcal{P}:a=t_0<t_1<\dots<t_m=b$ be a partition, and let $m_i$ and $M_i$ be the infimum and supremum of $h$ on $[t_{i-1},t_i]$, so that the lower and upper sums are $L(h,\mathcal{P})=\sum_i m_i(t_i-t_{i-1})$ and $U(h,\mathcal{P})=\sum_i M_i(t_i-t_{i-1})$. The nonnegative simple functions $\varphi=\sum_i m_i\mathbf{1}_{(t_{i-1},t_i]}$ and $\psi=\sum_i M_i\mathbf{1}_{(t_{i-1},t_i]}$ satisfy $\int\varphi\,d\lambda=L(h,\mathcal{P})$ and $\int\psi\,d\lambda=U(h,\mathcal{P})$: the displayed representations need not be the standard representations of Simple Function and Its Integral (several $i$ may share a value), but grouping the disjoint intervals $(t_{i-1},t_i]$ by value and using finite additivity of $\lambda$ (from Measure, Measure Space, and Probability Measure) together with interval lengths gives exactly these sums. Pointwise

since $m_i\le h\le M_i$ on $[t_{i-1},t_i]\supseteq(t_{i-1},t_i]$. Moreover $\int\tilde{h}\,\mathbf{1}_{(a,b]}\,d\lambda=\int\tilde{h}\,d\lambda$: the difference $\tilde{h}-\tilde{h}\mathbf{1}_{(a,b]}=h(a)\,\mathbf{1}_{\{a\}}$ is a simple function with integral $h(a)\,\lambda(\{a\})=0$, since the interval $[a,a]$ has length $0$; apply linearity. Hence by monotonicity (Linearity and Monotonicity of the Lebesgue Integral),

Step 4 (Darboux–Riemann bridge and conclusion). Let $\varepsilon>0$ and choose $\delta>0$ as in Riemann Integrability on a Closed Interval for $h$ and $\varepsilon$. By claim 2 of The Archimedean Property of the Real Numbers, applied with $x=b-a$ and $\varepsilon=\delta$ and writing $m$ for the image of the natural number it provides under the canonical map of $\mathbb{R}$, pick $m$ with $b-a<m\delta$, that is with $(b-a)/m<\delta$, and let $\mathcal{P}$ be the uniform partition $t_i=a+i(b-a)/m$, of mesh less than $\delta$. By the definition of the suprema $M_i$ there are tags $\xi_i\in[t_{i-1},t_i]$ with $h(\xi_i)>M_i-\varepsilon/(b-a)$; the corresponding Riemann sum $S^{+}=\sum_ih(\xi_i)(t_i-t_{i-1})$ satisfies $|S^{+}-I|<\varepsilon$ and

Symmetrically, by the definition of the infima $m_i$ there are tags $\eta_i$ with $h(\eta_i)<m_i+\varepsilon/(b-a)$, giving $L(h,\mathcal{P})>I-2\varepsilon$. Combining with Step 3 applied to this particular $\mathcal{P}$,

Since $\varepsilon>0$ was arbitrary, $\int_{\mathbb{R}}\tilde{h}\,d\lambda=I$. Undoing the shift of Step 2 gives Claim 3 in general. $\blacksquare$